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Maslowich
3 years ago
6

What is the frequency of a wave with a wavelength of 12 meters and a velocity of 4 m/s?

Physics
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

0.33 hz

Explanation:

the formula for the frequency in this situation is f=v/wavelength

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krek1111 [17]
It's average speed during that 26 seconds was about 4.77 m/s. Without seeing the graph, we can't tell if it was going faster or slower at any particular time during that period. All we can tell is its average for the full interval.
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Which statement is true? The speed of sound in air is inversely proportional to the temperature of the air. The speed of sound i
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6 0
3 years ago
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A car is traveling at a speed of 45 km/h into town. It takes the car 2 hours to get there. How far has the car traveled?
hjlf

Answer:

90 km

Explanation:

45 km per hr

2 hrs

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Hope this helps!

7 0
3 years ago
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sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he
podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0
3 years ago
Please help !! In the diagram, q1 = +0.00200 C, q2 = 0.00180 C, and q3 = +0.00830 C. the net force on q2 is zero. how far is q2
VikaD [51]

Answer:

2.03715

Explanation:

32364=8.99\cdot 10^9\cdot \frac{0.00180\cdot 0.00830}{2.03715^2}

4 0
3 years ago
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